2. BASIC MATH FOR RELAY TECHNICIANS
BONNEVILLE POWER ADMINISTRATION
for
WSU HANDS-ON RELAY SCHOOL

3. INTRODUCTION - Math I
This lecture will cover aspects of DC vs AC, why AC is so different, what is a sine wave and how it relates to phasors. We will also demonstrate how to use calculations involving right triangles to determine power factor, and to plot phasors for protective relay work.

4. In Service Lab
Thursday afternoon there will be an opportunity to explore the application of phasors as used in determining whether sensing devices are wired correctly, and reporting accurate information to relays.
The following information will help understand what is being done.

5. Mathematical Relationship of R, I, V, and P in DC
All you need to Remember is: V=IR (Ohms law), and P=VI (Joules law)… The rest just simple algebraic manipulation and substitution
[A] V=IR  [1] I= V/R and [2] R = V/I
[B] P=VI  [3] I= P/V and [4] V = P/I
P using [B] and [A]  P = (IR)I = I2R
P using [B] and [1]  P = V(V/R) = V2/R

6. AC vs DC Due to the nature of AC, not only do quantities such as voltage and current oscillate, but we have influences (inductance, capacitance) that can effect both a measured value and displacement in time.

7. AC generation as the magnet turns…
This way of creating energy that builds and collapses in different directions creates a sine wave…

8. The way AC is generated creates a ‘sine wave.’

9. AC Sinusoidal Wave Magnitude Relationships
VPEAK is VP-P / 2
In the power world, we refer to voltages at RMS value, not peak.

10. Example of two wave cycles
Example of two wave cycles. For relay work, this sinusoidal representation can be plotted on a coordinate system, a common language in the relay world for both engineers and craftspeople.
Remember: one cycle = one 360° rotation. It can be measure from any point on the sine wave, for example, from 270° to 270°.

11. One cycle of the same sine wave we just viewed can be graphically represented on a Cartesian coordinate system similar to this one, depending on where in time you take your measurements. One critical point, you must draw it with reference to it’s place in time (direction) and at it’s full RMS value.
90˚ (or -270)
Magnitude of 2 volts
Direction of 45˚
0, or 360˚
180˚
Plotted with Lead angles
270˚ (or -90)
RMS Value = 2 volts
45˚

13. Formulas for Z, I, V, P, Q, VA, PF, and RF in AC
Basic relationships
V=IZ, Volt-Amps=VI
Z = Impedance in ohms
Z=V/I*
(division reverses the polarity of the angle)
In addition:
Power Factor, or P.F. = COS 
Reactive Factor, or R.F. = SIN 
P = Watts = VA x P.F. = VI•COS 
Q = Vars = VA x R.F. = VI•SIN 
* = complex conjugate – division of polar coordinates – divide mags, subtract angles

14. Cartisian Coordinate System (used for plotting phasors)

15. We used this to plot points in school
We used this to plot points in school. For example: X = 6, Y = 3 (purple); or perhaps X = -10, Y = -2 (green) as seen below, to define a point on the coordinate system.

16. As an example, values as below can be representative of True Power (watts) on the horizontal line, and Reactive Power (vars) on the vertical line. By using those two quantities, we can find the ‘hypotenuse’ or VA.

17. Cartesian Coordinate System The summation of these quantities is proven through some basic trigonometric math. After we know the resultants magnitude and direction, we can then draw their resultant as a phasor.
y, or j, is 90˚
x, or r, is 0˚, (sometimes referred to as 360 ˚)
-x, or –r is 180˚
Phasors vs vectors
-y, or –j, is 270˚

18. Trigonometric Functions of a “Right Triangle”
Hypotenuse
Opposite
As viewed from this corner
 (angle of corner
Adjacent
Functions Inverse Functions
sine() = opp/hyp arcsine(opp/hyp) = SIN-1 = 
cos() = adj/hyp arccosine(adj/hyp) = COS–1 = 
tan() = opp/adj arctangent(opp/adj) = TAN-1 = 
It would be nice to have an easy way to remember this!

19. Trigonometric Functions of a “Right Triangle”
Sin = O oh
H heck
Cos = A another
H hour
Tan = O of
A Andy H O  A
Another way to remember is to use the mnemonic “Oh Heck Another Hour Of Andy!”
SOHCAHTOA
What you are doing is finding the ratio, or difference in size, between the numerator (top number) and the denominator (bottom number). This ratio will be given as a percentage. That percentage can then be used to find the angle of the triangle, or .
Example if H=10, O=5……
Sine = O/H = 5/10 = 0.5 (ratio), thereby SIN-1 of 0.5 = 30 degrees.

20. YOU DON’T NEED A FANCY CALCULATOR
YOU DON’T NEED A FANCY CALCULATOR. JUST SOMETHING SIMPLE SUCH AS THIS $14 TI-36X. YOU WILL NEED TRIGONOMETRIC FUNCTIONS (sine, cosine, tangent) AS WELL AS AN R»P / R«P FUNCTION (rectangular, polar).

21. Understanding the AC System using the Power Triangle Q: Why can’t AC power just be resistive? Q: What are Vars?
VA (Apparent Power)
VARS
(Reactive Power)
 Load Angle or power factor
WATTS (Real Power)

23. When inductors or capacitors are used in an AC circuit, the circuit is no longer purely resistive with voltage and current in phase. Rather, current and voltage do not cross the zero reference at the same time (shown below).
An inductor alone, without any circuit resistance, would cause current to lag by 90 degrees from its relative voltage. If resistance is added, that lag would decrease from 90 degrees depending on the amount of resistance.
A capacitor alone, without any circuit resistance, would cause current to lead its voltage by 90 degrees. If resistance is added, that lead would decrease from 90 degrees depending on the amount of resistance.
ELI the ICEman

24. Phasor Diagrams
The usual reference for zero phase is taken to be the positive x-axis and is associated with the resistor, or resistive part of an AC power system, since the voltage and current associated with the resistor are in phase.
The length of the phasor is proportional to the magnitude of the quantity represented, and its angle represents its phase relative to that of the current through the resistor. The phasor diagram for the RLC series circuit shows the main features.

25. Resistor AC Response
Impedance
                  
                                                                                       
Phasor diagram
                       
                                              

26. Capacitor AC Response Impedance Phasor diagram
                  
                                                                                       
                                                 
Phasor diagram
                   
ω = 2πf
You know that the voltage across a capacitor lags the current (ICE) because the current must flow to build up the charge, and the voltage is proportional to that charge which is built up on the capacitor plates.

27. Inductor AC Response Impedance Phasor diagram
                  
                                                                                       
Phasor diagram
                  
                                             
An induced electromotive force (emf) always gives rise to a current whose magnetic field opposes the change in original magnetic flux.
You know that the voltage across an inductor leads the current (ELI) because of Lenz' law. The inductor resists the buildup of the current, and it takes time for an imposed voltage to force the buildup of current to its maximum.

28. Q = VI sin , units Var (VA reactive)
Now that we understand the influence Vars have, let’s talk some more about triangles. Particularly, a power triangle.
Apparent Power
S = VI, units VA
Real Power
P = VI cos , units Watts
Reactive Power
Q = VI sin , units Var (VA reactive)
Lagging load (positive vars)
Leading load (negative vars)

29. Let’s use what we have learned so far to find VA of a circuit using watts and vars from board meters.
VA (Apparent Power) = ????
MVARS = 57.1 out
(Reactive Power)
 Load Angle or power factor = ????
WATTS = in (Real Power)

30. Which values do we know as given from the board meters?
Sin = O oh
H heck
Cos = A another
H hour
Tan = O of
A Andy
VA (Apparent Power) = ????
MVARS = 57.1 out
(Reactive Power)
 Load Angle or power factor = ????
WATTS = in (Real Power)
Which values do we know as given from the board meters?

31. So, which math function above would I use?
Sin = O oh
H heck
Cos = A another
H hour
Tan = O of
A Andy
( Hypotenuse) VA (Apparent Power) = ????
(Opposite) MVARS = 57.1 out
(Reactive Power)
 Load Angle or power factor = ????
(Adjacent) WATTS = in (Real Power)
So, which math function above would I use?

32. Since I have ‘adjacent’ & ‘opposite’ values, then use Tangent!
Sin = O oh
H heck
Cos = A another
H hour
Tan = O of
A Andy
( Hypotenuse) VA (Apparent Power) = ????
(Opposite) MVARS = 57.1 out
(Reactive Power)
 Load Angle or power factor = ????
(Adjacent) WATTS = in (Real Power)
Since I have ‘adjacent’ & ‘opposite’ values, then use Tangent!

33. Sin = O oh
H heck
Cos = A another
H hour
Tan = O of
A Andy
( Hypotenuse) VA (Apparent Power) = ????
(Opposite) MVARS = 57.1 out
(Reactive Power)
 Load Angle or power factor = ????
(Adjacent) WATTS = in (Real Power)
Answer: Tan = O/A; Tan = 57.1/640.2; = ; ArcTan of = ˚ Our angle is now known.

34. Sin = O oh
H heck
Cos = A another
H hour
Tan = O of
A Andy
Now, let’s use the known angle of ˚ to see what VA, or the length of the hypotenuse line, is?
(Opposite) MVARS = 57.1 out
(Reactive Power)
( Hypotenuse) VA (Apparent Power) = ????
 Load Angle or power factor = ????
(Adjacent) WATTS = in (Real Power)
For this, now knowing three quantities, you can use any of the equations. Let’s use Sin! Sin = O/H Sin of angle ˚ = / H; = 57.1 / H; H = 57.1/ ; H = VA! To prove this answer, we can use Pythagoreans theorem, which states C² = A² + B², or C = √(A² + B²) (SEE NEXT SLIDE)

35. Sin = O oh
H heck
Cos = A another
H hour
Tan = O of
A Andy
To prove this answer, we can use Pythagoreans theorem, which states C² = A² + B², or C = √(A² + B²) Therefore: = √(640.2² ²). Did we prove our math???
(Opposite) MVARS = 57.1 out
(Reactive Power)
( Hypotenuse) VA (Apparent Power) = 642.7 C B
 Angle = A
(Adjacent) WATTS = in (Real Power)

36. Q: why is the triangle upside down?
Now, let’s plot our known quantities on a Cartesian coordinate system.
We know that we have:
57.1 Vars out
640.2 watts in
An angle of degrees
VA = (basically IE)
Q: why is the triangle upside down?
A: LEAD VS LAGGING
Additionally, we also know that VA = IE. If voltage is 289V,
and we decide to plot that voltage at 0 degrees just like resistance/watts, that means current (I) would lag its respective voltage by the same angle used in finding VA.

37. The previous triangle was plotted upside down because we have ‘vars out’ according to our board. Which means our load on the end of the line looks like what? A capacitor or Inductor? What do you think?

38. Now, let’s plot our known quantities on a Cartesian coordinate system.
We know that we have:
57.1 Vars out
640.2 watts in
An angle of degrees
VA = (basically IE) I = 669 amps
V = 500KV/√3 = 289KV
Additionally, we also know that VA = IE. If voltage is 289V,
and we decide to plot that voltage at 0 degrees just like resistance/watts, that means current (I) would lag its respective voltage by the same angle used in finding VA. Congratulations! We just plotted phasors! 

39. REAL AND REACTIVE POWER AT 70% P.F.
Sinusoidal “graphical” Comparison of V & I in AC
REAL AND REACTIVE POWER AT 70% P.F.
Cos of 0 = 1,
Therefore PF = 1
Cos (PF) ≈ 45˚
REAL POWER ONLY

40. Graphic Vector Addition
Adding and subtracting (graphically)
ADDING – TAIL to HEAD
SUBTRACTING – still TAIL to HEAD (after subtracted vector is rotated 180°)

41. Rectangular Coordinates
As we have seen, there are polar values, or rectangular values. Both show the same result, but polar is used when graphing phasors on a coordinate system.
Polar Coordinates
Rectangular Coordinates

42. Rectangular Coordinate Math
Adding & Subtracting (easy)
Multiplication (hard)
Division (hardest)
Adding & Subtracting

43. Polar Coordinate Math
Multiplication (easy)
Division (easy)

44. Polar to Rectangular and Rectangular to Polar Conversions
P  R
x = r cosq   y = r sinq
R  P
q = tan-1( y/x )

45. Polar to Rectangular and Rectangular to Polar Conversions
Calculator that is capable of performing complex math or
Use PR / RP conversion calculator functions
Adding and subtracting (using PR conversion)
Multiplication & Division (using RP conversion)
Convert back and forth as need based on whether the
operation is Addition / Subtraction or
Multiplication / Division

46. Example
Verify the line current based on the two ring bus breaker currents:
50a < -150 and 70a < -300

47. Example Answer Enter 50, x><y, -15
2nd button, P>R, jot down value, then x><y and jot down second value (48.3, -12.9)
Do this again for 70<-30
Add the X values and add the Y values
Enter the resulting X, x><y, resulting Y, 3rd, jot down value, x><y, jot down second value
118a<-23.70

48. A Variety of Planes
Voltage and current phasors are represented on a plane that represents these magnitudes
We use separate planes to represent other quantities
P = VI so the power phasor has the same angle direction as the current phasor
A bit confusing but
lagging current yields what
are called “positive” Vars

49. A Variety of Planes
Z = V/I so the +/- sign of the angle changes between injection plane and impedance plane
If the impedance is at 800, the current injection will be at – 800
The plane is defined by resistance and reactance

50. A Variety of Planes
Transformer differential is represented on yet another plane with restraint and operate currents defining the plane

51. Ratios Ratios are used to compare two quantities
They can be expressed as comparisons - 2:3 or as fractions - 2/3
We see them in transformers – 115kV/12.47kV
The full description of transformer parameters is:
a (turns ration) = N1 / N2 = V1 / V2 = I2 / I1
Can be simplified by dividing 115/12.47 = 9.2/1
Also CTs 2000/5 = 400/1
And PTs 115,000/115 = 1000/1

52. Per Unit Per unit means putting the base at one
Like percentage without the factor of 100
50% is the same as 0.5 per unit
A relay value called tap is a base – TOC
2 x tap = 2 per unit
Used in some relay settings – differential minimum op and unrestrained are often per unit settings, also directional power in some generator relays

53. Per Unit Example if HS = 230 kV it would be 2VbaseHS
40MVA transformer, 115 kV / 12.47kV
40MVA = Sbase, 115kV = VbaseHS, 12.47kV = Vbase LS
Rated current would be
40,000,000va/115,000 v/ √3 = a = IbaseHS
If HS = 100 kV, this would be 0.87 VbaseHS
(100kV/115kV = 0.87pu)
if HS = 230 kV it would be 2VbaseHS
(230kV/115kV = 2.0pu)
If you measure a, this would be 0.5 IbaseHS
(301.2a/602.4a = 0.5pu)
If you run it at 110 kV and 50 a, you would see 110 x 50 x √3 = 9.53 MVA or about 0.24 Sbase
(9.53MVA/40MVA = 0.24pu)

54. Thank you!!!

55. Transformer Ratio relationships by Turns, Voltage, and Current
Turns ratio = Number of primary turns / Number of secondary turns
V ratio = Turns ratio = VPRIMARY / VSECONDARY
I ratio = 1 / Turns Ratio = VSECONDARY / VPRIMARY = IPRI / ISEC

56. Instantaneous “phasor” Polarity of current and voltage on a transformer
Voltage Aligns With Polarity on Both Windings
Current In Polarity Terminal of One Winding Comes Out the Polarity Terminal of the Other Winding